# Comments are in maroon Code is in black Results are in this green## rep()

# Often we want to start with a vector of 0's and then modify the entries in later code. R makes this easy with the replicate function rep() # rep(0, 10) makes a vector of of 10 zeros. x = rep(0,10) x [1] 0 0 0 0 0 0 0 0 0 0 # rep() will replicate almost anything x = rep(2,6) x [1] 2 2 2 2 2 2 x = rep('abc',5) x [1] "abc" "abc" "abc" "abc" "abc" x = rep(1:4,5) x [1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4## for(i in x)

# 'for loops' let us repeat (loop) through the elements in a vector and run the same code on each element # We will illustrate with examples # Loop through the sequence 1 to 5 printing the square of each number for (j in 1:5) { print(j^2) } class='r'>[1] 1 [1] 4 [1] 9 [1] 16 [1] 25 # We can capture the results of our loop in a list # First we create a vector and then we fill in its values n = 5 x = rep(0,n) for (j in 1:n) { x[j] = j^2 } x class='r'>[1] 1 4 9 16 25 # You always wanted to know the sum of the first 100 squares. n = 100 x = rep(0,n) for (j in 1:n) { x[j] = j^2 } sum(x) class='r'>[1] 338350 # Let's use a for loop to estimate the average of squaring the result of a roll of a die. nsides = 6 ntrials = 1000 trials = rep(0,ntrials) for (j in 1:ntrials) { trials[j] = sample(1:nsides,1) # We get one sample at a time } mean(trials^2) [1] 15.207 # Of course we could have done this simulation without a loop. # for loops are truly valuable when the calculation is more complicated and we can't do it exactly or with built in R functions. # Let's estimate the probability of a derangement in a permutation of 9 objects. (A derangement is a permutation where no element ends up in its original position.) n = 9 x = 1:n ntrials = 10000 trials = rep(0,ntrials) for (j in 1:ntrials) { y = sample(x,n) s = sum(y == x) # s = number of people in their original seat trials[j] = (s == 0) # 1 if a derangement, 0 if not } mean(trials) # mean(trials) = fraction that are 1's [1] 0.3697